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Thread: Log Return Question

  1. #1
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    Log Return Question

    New to the forum. Math has been a lifetime of frustration for me so apologies if I don't articulate this question using the appropriate terminology. I know that if you take the ln(x) - ln(x-1) this is the log return. I think 1 + this log return would be considered comparable to a simple return of x / x-1. The question I came here for is what does ln(x) / ln(x-1) tell you??

    Thanks for your time and help.
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  2. #2
    Senior Member x3bnm's Avatar
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    Re: Log Return Question

    $\displaystyle \ln(x) - \ln(x-1) $


    $\displaystyle =\ln(x) - \ln(x-1)$

    $\displaystyle =\ln\left(\frac{x}{x-1}\right) $






    $\displaystyle 1 + \log\left(\ln\left(\frac{x}{x - 1}\right) \right) \ne \frac{x}{x-1}$

    You can check what I just stated here. Wolfram|Alpha: Computational Intelligence

    And $\displaystyle \frac{\ln(x)}{\ln(x-1)} = \frac{\text{log of x base e}}{\text{log of (x - 1) base e}}$

    Also $\displaystyle \frac{\ln(x)}{\ln(x-1)} = \log_{(x-1)} {x}$

    http://www.wolframalpha.com/input/?i...+log_(x-1)+(x)

    I hope it helps.
    Last edited by x3bnm; Aug 1st 2018 at 04:22 AM.
    Thanks from topsquark
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